📜  祖克曼数字

📅  最后修改于: 2021-04-24 19:22:29             🧑  作者: Mango

给定整数N ,任务是检查N是否为Zuckerman数。

例子:

方法:想法是找到N的数字乘积,并检查N是否可被其数字的乘积整除。如果是,则数字N是祖克曼数字。
下面是上述方法的实现:

C++
// C++ implementation to check if N
// is a Zuckerman number
 
#include 
using namespace std;
 
// Function to get product of digits
int getProduct(int n)
{
    int product = 1;
 
    while (n != 0) {
        product = product * (n % 10);
        n = n / 10;
    }
 
    return product;
}
 
// Function to check if N is an
// Zuckerman number
bool isZuckerman(int n)
{
    return n % getProduct(n) == 0;
}
 
// Driver code
int main()
{
    int n = 115;
    if (isZuckerman(n))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}


Java
// Java implementation to check if N
// is a Zuckerman number
class GFG{
 
// Function to get product of digits
static int getProduct(int n)
{
    int product = 1;
     
    while (n != 0)
    {
        product = product * (n % 10);
        n = n / 10;
    }
    return product;
}
 
// Function to check if N is an
// Zuckerman number
static boolean isZuckerman(int n)
{
    return n % getProduct(n) == 0;
}
 
// Driver code
public static void main(String[] args)
{
    int n = 115;
     
    if (isZuckerman(n))
    {
        System.out.println("Yes");
    }
    else
    {
        System.out.println("No");
    }
}
}
 
// This code is contributed by shubham


Python 3
# Python3 implementation to check if N
# is a Zuckerman number
 
# Function to get product of digits
def getProduct(n):
    product = 1
 
    while (n > 0):
        product = product * (n % 10)
        n = n // 10
 
    return product
 
# Function to check if N is an
# Zuckerman number
def isZuckerman(n):
 
    return n % getProduct(n) == 0
 
# Driver code
N = 115
if (isZuckerman(N)):
    print("Yes")
else:
    print("No")
     
# This code is contributed by Vishal Maurya


C#
// C# implementation to check if N
// is a Zuckerman number
using System;
class GFG{
  
// Function to get product of digits
static int getProduct(int n)
{
    int product = 1;
      
    while (n != 0)
    {
        product = product * (n % 10);
        n = n / 10;
    }
    return product;
}
  
// Function to check if N is an
// Zuckerman number
static bool isZuckerman(int n)
{
    return n % getProduct(n) == 0;
}
  
// Driver code
public static void Main(String[] args)
{
    int n = 115;
      
    if (isZuckerman(n))
    {
        Console.WriteLine("Yes");
    }
    else
    {
        Console.WriteLine("No");
    }
}
}
 
// This code is contributed by Rohit_ranjan


Javascript


输出
Yes

参考文献: OEIS